Acid-Base Titrations

Transcription

1 Acid-Base Titrations

2 The Titration One of the most important lab procedures involving acids and bases is the titration. A titration is an analytical procedure that allows for the measurement of the amount of one solution that is required to exactly react with the contents of another solution. In acid-base terms, you add one solution to the other until the equivalence point is reached. The use of a ph meter will produce a ph curve (titration curve), so you can specifically calculate at what ph your solutions have been neutralized.

3 Titration Types 1. acid-base HCl + NaOH H 2 O + NaCl 2. redox 5NaC 2 O KMnO 4 + 6H + 10CO 2 + 2Mn 2+ +8H 2 O 3. precipitation Ag + + X - AgX(s) 4. complex formation (EDTA) ethylenediaminetetraacetic acid

4 Equivalence Point The point at which there are stoichiometric equivalent amounts of a cid and base. [H + ] = [OH - ] Determined by indicator color change dramatically change in ph

5 Acid-base indicators General Many indicators are weak acids and partially dissociate in aqueous solution HIn (aq) H + (aq) + In (aq) The un-ionised form (HIn) is a different colour to the anionic form (In ).

6 Acid-base indicators General Many indicators are weak acids and partially dissociate in aqueous solution HIn (aq) H + (aq) + In (aq) The un-ionised form (HIn) is a different colour to the anionic form (In ). Apply Le Chatelier s Principle to predict any colour change In acid - increase of [H + ] - equilibrium moves to the left to give red undissociated form In alkali - increase of [OH ] - OH ions remove H + ions to form water; H + (aq) + OH (aq) H 2 O(l) - equilibrium will move to the right to produce a blue colour

7 Acid-base indicators General Many indicators are weak acids and partially dissociate in aqueous solution HIn (aq) H + (aq) + In (aq) The un-ionised form (HIn) is a different colour to the anionic form (In ). Apply Le Chatelier s Principle to predict any colour change In acid - increase of [H + ] - equilibrium moves to the left to give red undissociated form In alkali - increase of [OH ] - OH ions remove H + ions to form water; H + (aq) + OH (aq) H 2 O(l) - equilibrium will move to the right to produce a blue colour Summary In acidic solution HIn (aq) H + (aq) + In (aq) In alkaline solution

8 Acid-base indicators Must have an easily observed colour change. Must change immediately in the required ph range over the addition of half a drop of reagent. COLOUR CHANGES OF SOME COMMON INDICATORS ph METHYL ORANGE LITMUS PHENOLPHTHALEIN CHANGE CHANGE CHANGE

9 The ph range of indictors Indictors dose not change colour sharply at one particular ph, they change over a narrow range of ph indctors pkind ph litmus methylorange phenophthaline

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11 Acid-base indicators Must have an easily observed colour change. Must change immediately in the required ph range over the addition of half a drop of reagent. To be useful, an indicator must change over the vertical section of the curve where there is a large change in ph for the addition of a very small volume of alkali. The indicator used depends on the ph changes around the end point - the indicator must change during the vertical portion of the curve. PHENOLPHTHALEIN LITMUS METHYL ORANGE In the example, the only suitable indicator is PHENOLPHTHALEIN.

12 ph curves Types There are four types of acid-base titration; each has a characteristic curve. strong acid (HCl) v. strong base (NaOH) weak acid (CH 3 COOH) v. strong alkali (NaOH) strong acid (HCl) v. weak base (NH 3 ) weak acid (CH 3 COOH) v. weak base (NH 3 ) In the following examples, alkali (0.1M) is added to 25cm 3 of acid (0.1M) End points need not be neutral due to the phenomenon of salt hydrolysis

13 Strong Acid-Strong Base This diagram shows the change in ph as a solution of strong acid is slowly added to a solution of strong base. Strong acid-strong base titrations usually have an equivalence point around 7. This is due to the fact that in solution, strong acids and strong bases will both completely dissociate, so there is an approximate 1:1 ratio of moles of acid to moles of base at the equivalence point.

14 strong acid (HCl) v. strong base (NaOH)

15 ph 1 at the start due to 0.1M HCl (strong monoprotic acid) strong acid (HCl) v. strong base (NaOH)

16 strong acid (HCl) v. strong base (NaOH) ph 1 at the start due to 0.1M HCl (strong monoprotic acid) Very little ph change during the initial 20cm 3

17 strong acid (HCl) v. strong base (NaOH) Very sharp change in ph over the addition of less than half a drop of NaOH ph 1 at the start due to 0.1M HCl (strong monoprotic acid) Very little ph change during the initial 20cm 3

18 strong acid (HCl) v. strong base (NaOH) Curve levels off at ph 13 due to excess 0.1M NaOH (a strong alkali) Very sharp change in ph over the addition of less than half a drop of NaOH ph 1 at the start due to 0.1M HCl (strong monoprotic acid) Very little ph change during the initial 20cm 3

19 strong acid (HCl) v. strong base (NaOH) PHENOLPHTHALEIN LITMUS METHYL ORANGE Any of the indicators listed will be suitable - they all change in the vertical portion

20 strong acid (HCl) v. weak base (NH 3 ) Curve levels off at ph 10 due to excess 0.1M NH 3 (a weak alkali) Sharp change in ph over the addition of less than half a drop of NH 3 ph 1 at the start due to 0.1M HCl Very little ph change during the initial 20cm 3

21 strong acid (HCl) v. weak base (NH 3 ) PHENOLPHTHALEIN LITMUS METHYL ORANGE Only methyl orange is suitable - it is the only one to change in the vertical portion

22 Weak Acid-Strong Base At the beginning, the weak acid is in equilibrium with itself. Thus, the beginning ph is higher than that of a strong acid. More moles of the weak acid are required to completely react with the amount of strong base present, because the acid is not entirely dissociated. For every mole of base, there is less than one mole of acid able to react: there is an excess of base. The initial change in ph is steeper because there is excess base. Weak acid-strong base titrations reach their equivalence point at a ph greater than 7.

23 weak acid (CH 3 COOH) v. strong base (NaOH) Curve levels off at ph 13 due to excess 0.1M NaOH (a strong alkali) Sharp change in ph over the addition of less than half a drop of NaOH Steady ph change ph 4 due to 0.1M CH 3 COOH (weak monoprotic acid)

24 weak acid (CH 3 COOH) v. strong base (NaOH) PHENOLPHTHALEIN LITMUS METHYL ORANGE Only phenolphthalein is suitable - it is the only one to change in the vertical portion

25 weak acid (CH 3 COOH) v. weak base (NH 3 ) Weak acid-weak base titrations are difficult because each solution is in equilibrium. The ph of the equivalence point is around 7.

26 weak acid (CH 3 COOH) v. weak base (NH 3 ) Curve levels off at ph 10 due to excess 0.1M NH 3 (a weak alkali) Steady ph change NO SHARP CHANGE IN ph ph 4 due to 0.1M CH 3 COOH (weak monoprotic acid) Types

27 weak acid (CH 3 COOH) v. weak base (NH 3 ) PHENOLPHTHALEIN LITMUS METHYL ORANGE NOTHING SUITABLE There is no suitable indicator- none change in the vertical portion. The end point can be detected by plotting a curve using a ph meter.

28 Ex. 1 What is the concentration of HCl if 30.0 ml of 0.10 M NaOH neutralizes 50.0mL HCl? NaOH + HCl H 2 O Use M a = nm a V a =nm b V b + NaCl How many moles of HCl were used? moles= M a V a, but convert the volume to L( 50mL=0.05L).

29 Titration 42.5 ml of 1.3M KOH are required to neutralize 50.0 ml of H 2 SO 4. Find the molarity of H 2 SO 4. H 3 O + M =? V = 50.0 ml n = 2 OH - M = 1.3M V = 42.5 ml n = 1 nmv =n MV M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H 2 SO 4 Courtesy Christy Johannesson

30 Sample Calculation: Strong Acid-Strong Base Titration Curve erive the curve of the titration of 40.0 ml of M HCl with M NaOH. Region 1. Before the equivalence point, after adding 20.0 ml of M NaOH. (Half way to the equivalence point.) Initial mmoles of H 3 O + = 0.100* 40 ml = 4 mmoles of OH - added =0.1*20 ml= 2 amount (mmol)of H3O remaining [H3O ] original volumeof acid volumeof added base = (4-2)/60 = ph = - log [H + ] = - log = 1.48

31 Sample Calculation: Strong Acid-Strong Base Titration Curve Region 2. At the equivalence point, after adding 40.0 ml of M NaOH. Initial mmoles of H 3 O + 40 ml x M = 4 mmol H 3 O + mmoles of OH - added 40ml x M =4 mmol OH - OH H K w ph =7

32 Sample Calculation: Strong Acid-Strong Base Titration Curve Region 3. After the equivalence point, after adding 50.0 ml of M NaOH. (Now calculate excess OH - ) Total mmoles of OH - = 50 ml x M = 5 mmol OH - mmoles of H 3 O + consumed = 40 ml x M =4 mmol [OH ] =5-4/90 =0.011 poh amount (mmol)of OH remaining original volumeof acid volumeof added base log OH =1.95 ph = =12.05

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34 Titration of strong base with strong acid In the titration of any strong base with any strong acid, H + + OH - H 2 O there are three regions of the titration curve that require different kinds of calculations: 1. Before the equivalence point, the ph is determined by excess OH in the solution. poh log OH 2. At the equivalence point, H + is just sufficient to react with all OH to make H 2 O. The ph is determined by dissociation of water. OH H K w 3. After the equivalence point, ph is determined by excess H + in the solution. ph log H

35 Calculating the ph during a weak acid strong base titration [H 3 O + ][A - ] K a = [HA] ph = -log[h 3 O + ] Initial ph [H 3 O + ] = ph before equivalence point [H 3 O + ] = K a x [HA] or [A - ] ph = pk a + log [base] [acid]

36 Calculating the ph during a weak acid strong base titration ph at the equivalence point A - (aq) + H 2 O(l) HA(aq) + OH - (aq) [OH - ] = where [A - ] = mol HA init and K b = K w V acid + V base K a K [H 3 O + ] w and ph = -log[h 3 O + ] ph beyond the equivalence point [OH - ] = mol OH - excess V acid + V base [H 3 O + ] = ph = -log[h 3 O + ] K w [OH - ]

37 Titration of a Weak Acid by a Strong Base Derive the curve of the titration of 25.0 ml of M HA (acetic acid) is titrated with M NaOH. (K a = 1.8 x 10 5 ) Determine ph in each of four regions: (a) at the start -- before any NaOH is added Just a simple weak acid: ph = 2.72 H C HA K a

38 (b) before the equivalence point: after addition of 10.0 ml of NaOH The acid is partially neutralized by the added OH HA + OH --> H 2 O + A 25x0.2 10x0.2 10x0.2 Since both HA and A are now present, the mixture is a buffer solution! ph = pk a + log (C A- /C HA ) [HA] left = ( )/35 = [A - ] produced = (10 0.2)/35 = ph = (-log ) + log (0.057/0.085) ph = 4.57

39 (c) at the equivalence point: The acid (HA) is completely neutralized, and only the anion A is present the solution is now the salt of a weak acid! Total volume is now 50.0 ml, so [A ] = 0.10 M and ph = 8.88 (d) after the equivalence point: All acid (HA) is gone and excess OH is present the ph is determined by how much OH is left after all the acid has been neutralized e.g. after 40.0 ml of NaOH added: ph = 12.66

40 Titration Curve

41 Finding the ph During a Weak Acid Strong Base Titration PROBLEM: Calculate the ph during the titration of ml of M propanoic acid (HPr; K a = 1.3x10-5 ) after adding the following volumes of M NaOH: (a) 0.00 ml (b) ml (c) ml (d) ml PLAN: The initial ph must be calculated using the K a value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH - added. Once we know the volume of base required to reach the equivalence point we can calculate the ph based on the species present in solution. SOLUTION: (a) [H 3 O + ] = = 1.1x10-3 M ph = -log(1.1x10-3 ) = 2.96

42 (b) ml of M NaOH has been added. Initial amount of HPr = L x M = 4.000x10-3 mol HPr Amount of NaOH added = L x M = 3.000x10-3 mol OH - Each mol of OH - reacts to form 1 mol of Pr -, so Concentration (M) HPr(aq) + OH - (aq) Pr - (aq) + H 2 O(l) Initial Change Equilibrium [H 3 O + ] = K a x [HPr] [Pr - ] ph = -log(4.3x10-6 ) = 5.37 = (1.3x10-5 ) x = 4.3x10-6 M

43 (c) ml of M NaOH has been added. This is the equivalence point because mol of OH - added = = mol of HA init. All the OH - added has reacted with HA to form mol of Pr -. [Pr - ] = mol L L = M Pr - is a weak base, so we calculate K b = [H 3 O + ] K w = K w K a 1.0x10-14 = 1.6x10-9 M = 1.0x x10-5 = 7.7x10-10 ph = -log(1.6x10-9 ) = 8.80

44 (d) ml of M NaOH has been added. Amount of OH - added = L x M = mol Excess OH - = OH - added HA init = = mol [OH - ] = mol OH- excess total volume = mol L = M [H 3 O + ] = K w [OH - ] = 1x = 9.0x10-13 M ph = -log(9.0x10-13 ) = 12.05

45 The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve 1. The initial ph is higher. 2. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. 3. The ph at the equivalence point is greater than The steep rise interval is less pronounced.

46 TITRATION CURVES II. Weak acid with strong base Weak base with strong acid e.g. Titration of CH 3 COOH with NaOH, Titration of NH 4 OH with HCl: Weak acid I. At the start: ph Weak base H K a Cacid K CH COOH H a 3 OH K b C base K NH OH OH b 4 II. Before the end point: Buffer (acid / salt) ph Buffer (base / salt) Cacid K Cbase OH K a a CH 3COOH H b OH CH COO H K C salt 3 K NH OH III. At the end point: Hydrolysing salt (Brönsted base) ph Hydrolysing salt (Brönsted acid) K OH w Csalt K H w Csalt K OH K b Csalt a K CH COO OH b 3 IV. After the end point: Excess of strong base ph Excess of strong acid [OH ] = C excess base [OH ] = [NaOH] excess [H + ] = C excess acid [H + ] = [HCl excess K C salt b NH 4 4 b H K a NH 4 H K a Csalt

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